The Ballet of Triangle Centers on the Elliptic Billiard

Journal for Geometry and Graphics Volume 24 (2020), No. 1, 79–101.

Dan S. Reznik1, Ronaldo Garcia2, Jair Koiller3 1Data Science Consulting Rio de Janeiro RJ 22210-080, Brazil email: [email protected] 2Instituto de Matemática e Estatística, Universidade Federal de Goiás, Goiânia GO, Brazil email: [email protected] 3Universidade Federal de Juiz de Fora, Juiz de Fora MG, Brazil email: [email protected]

Abstract. We explore a bevy of new phenomena displayed by 3-periodics in the Elliptic Billiard, including (i) their dynamic geometry and (ii) the non-monotonic motion of certain Triangle Centers constrained to the Billiard boundary. Fasci- nating is the joint motion of certain pairs of centers, whose many stops-and-gos are akin to a Ballet. Key Words: elliptic billiard, periodic trajectories, triangle center, loci, dynamic geometry. MSC 2010: 51N20, 51M04, 51-04, 37-04

1. Introduction

We investigate properties of the 1d family of 3-periodics in the Elliptic Billiard (EB). Being uniquely integrable [13], this object is the avis rara of planar Billiards. As a special case of Poncelet’s Porism [4], it is associated with a 1d family of N-periodics tangent to a confocal Caustic and of constant perimeter [27]. Its plethora of mystifying properties has been extensively studied, see [26, 10] for recent treatments. Initially we explored the loci of Triangle Centers (TCs): e.g., the Incenter X1, Barycenter X2, Circumcenter X3, etc., see summary below. The Xi notation is after Kimberling’s Encyclopedia [14], where thousands of TCs are catalogued. Here we explore a bevy of curious behaviors displayed by the family of 3-periodics, roughly divided into two categories (with an intermezzo): 1. The shape of 3-periodics and/or TC loci: when are the former acute, obtuse, pythagorean, and the latter non-smooth, self-intersecting, non-compact? See for example this recent treatment of TCs at inﬁnity [15].

ISSN 1433-8157/$ 2.50 c 2020 Heldermann Verlag

80 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 2. The kinematics of EB-railed TCs: a handful of TCs1 is known to lie on the EB. As the family of 3-periodics is traversed, what is the nature of their motion (monotonicity, critical points, etc.). We further examine the joint motion of two EB-railed TCs which in some cases resembles a Ballet. In the intermezzo we introduce (i) a deceptively simple construction under which the EB bugles out the Golden Ratio, and (ii) a triangle closely related to 3-periodics2, whose vertices are dynamically clutched onto the EB. Throughout the paper ﬁgures will sometimes include a hyperlink to an animation in the format [23, PL#nn], where nn is the entry into our video list on Table 5 in Section 5. Indeed, the beauty of most phenomena remain ellusive unless they are observed dynamically.

Related Work Properties of the poristic triangle family, i.e., one with ﬁxed incircle and circumcircle, were studied in [29, 31, 18], where the loci of many triangle centers is described. Indeed, we noticed poristic triangles are related to the 3-periodic family by a variable similarity transform, therefore sharing with it all scale-free invariants [7]. Regarding 3-periodics, two intriguing invariants are [24]:

1. The Mittenpunkt X9 is the black swan of all TCs: its locus is a point at the center of the EB3 [23, PL#01]. 2. The 3-periodic family conserves the ratio of Inradius-to-Circumradius. This implies the sum of its cosines is invariant. Suprisingly the latter holds for all N-periodics [1, 2]. Other observations focused on the surprising elliptic locus of many TCs: the locus of the Incenter X1 is an ellipse [25], as is that of the Excenters [6]. The latter is similar to a rotated locus of X1, see [23, PL#01,04]. Also elliptic are the loci of the Barycenter X2 [28], Circumcenter X3 [5], Orthocenter X4 [6], Center X5 of the Nine-Point Circle [8], see [23, PL#05,07]. Recently we showed that out of the ﬁrst 100 Kimberling Centers in [14], exactly 29 produce elliptic loci [8]. This is quite unexpected given the non-linearities involved. Other observations, many which ﬁnd parallels in Triangle Geometry, included (i) the locus 4 of the Feuerbach Point X11 is on the Caustic , as is that of the Extouchpoints, though these move in opposite directions; (ii) the locus of X100, the anticomplement of X11, is on the EB [9]; (iii) the locus of vertices of Intouch, Medial and Feuerbach Triangles are all non-elliptic. We still don’t understand how loci ellipticity or many of the phenomena below correlate to the Trilinears of a given TC.

2. First movement: shape

Let the boundary of the EB be given by, a>b> 0:

x 2 y 2 f(x, y)= + =1 (1) a b Below c2 = a2 b2, and δ = √a4 a2b2 + b4. − − 1 Some 50 out of 40 thousand in [14]. 2 The Contact (or Intouch) Triangle of the Anticomplementary Triangle (ACT). 3 In Triangle parlance, the EB is the “X9-Centered Circumellipse”. 4 The Inconic centered on the Mittenpunkt X9 which passes through the Extouchpoints is known as the Mandart Inellipse [30]. By deﬁnition, an Inconic is internally tangent to the sides, so it must be the Caustic. D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 81

Figure 1: Let α4 = 2 √2 1 1.352 and H be the elliptic locus of X4 (orange), similar to a − ≃ rotated copy of the EBp (black). Top Left: a/b < α4, H is interior to the EB and all 3-periodics (blue) are acute. Bot Left: at a/b = α4, H is tangent to the top and bottom vertices of the EB. The 3-periodic shown is a right triangle since one vertex is at the upper EB vertex where X4 currently is. Right: at a/b > α4, the 3-periodic family will contain both acute and obtuse 3-periodics, corresponding to X4 interior (resp. exterior) to the EB. For any obtuse 3-periodic, X4 will lie within the wedge between sides incident upon the obtuse angle and exterior to the 3-periodic, i.e., exterior to the EB. For the particular aspect ratio shown (a/b = 1.51), H is identical to a 90◦-rotated copy of the EB.

Throughout this paper we assume one vertex P1(t) = (x1,y1) of a 3-periodic is 5 parametrized as P1(t) = [a cos t, b sin t]. Explicit expressions for the locations of P2(t) and P3(t) under this speciﬁc parametrization are given in Appendix B.

2.1. Can 3-periodics be obtuse? Pythagorean?

The locus of the Orthocenter X4 is an ellipse of axes a4, b4 similar to a rotated copy of the EB. These are given by [8]:

2 2 2 2 k4 k4 (a + b )δ 2 a b (a4, b4)= , , k4 = − a b c2

Referring to Figure 1, let α4 = 2 √2 1 1.352. p − ≃ Proposition 1. If a/b = α4, then b4 = b, i.e., the top and bottom vertices of the locus of X4 coincide with the EB’s top and bottom vertices.

5 Their coordinates involve double square roots on x1 so these points can be constructed by ruler and compass. 82 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard

4 2 2 4 Proof. The equation b4 = b is equivalent to a +2a b 7b =0. Therefore, as a>b> 0, it − follows that a/b = 2 √2 1. − ∗ p 6 4 3 2 ∗ Let α4 be the positive root of x + x 4 x x 1=0, i.e., α4 = 1.51. − − − ≃ ∗ Proposition 2. When a/b = α4, then a4 = b and b4 = a, i.e., the locus of X4 is identical to a rotated copy of Billiard.

6 4 2 3 3 2 4 6 Proof. The condition a4 = b, or equivalently b4 = a, is deﬁned by a +a b 4a b a b b = 6 4 3 2 − − − 0. Graphic analysis shows that x + x 4 x x 1=0 has only one positive real root ∗ − − − which we call α4.

Theorem 1. If a/b < α4 (resp. a/b > α4) the 3-periodic family will not (resp. will) contain obtuse triangles.

Proof. If the 3-periodic is acute, X4 is in its interior, therefore also internal to the EB. If the 3-periodic is a right triangle, X4 lies on the right-angle vertex and is therefore on the EB. If the 3-periodic is obtuse, X4 lies on exterior wedge between sides incident on the obtuse vertex (feet of altitudes are exterior). Since the latter is on the EB, X4 is exterior to the EB.

2.2. Can a locus be non-smooth? Loci considered thus far have been smooth, regular curves.

Proposition 3. If a/b > α4 the locus of the Incenter of the 3-periodic’s Orthic Triangle comprises four arcs of ellipses, connected at four corners.

6 To see this, let T be a triangle, Th its Orthic , and Ih be the latter’s Incenter. It is well-known that if T is acute Ih coincides with T ’s Orthocenter X4. However, for obtuse T :

Lemma 1. Th has two vertices outside of T , and Ih is “pinned” to the obtuse vertex. This is a known result [3, Chapter 1], which we revisit in Appendix A. This curious phenomenon is illustrated in Figure 2. Assume a/b > α4. Since the 3-periodic family contains both acute and obtuse triangles, the locus of Ih transition between acute and obtuse regimes:

3-periodic X4 wrt EB Ih location acute interior Orthocenter X4 right triangle on it right-angle vertex obtuse external (3-periodic Excenter) obtuse vertex, on EB

In turn, this yields a locus for Ih consisting of four elliptic arcs connected by their endpoints in four corners (Figure 3). Notice top and bottom (resp. left and right) arcs belong to the EB (resp. X4 locus). 2 ′2 For the next proposition, let αh (resp. 1/αh ) be the real root above 1 (resp. less than 1) of the polynomial 1+12x 122x2 52x3 + 97x4. Numerically, α 1.174 and α′ 2.605. − − h ≃ h ≃ ′ Proposition 4. At a/b = αh (resp. a/b = αh), at the sideways (resp. upright) 3-periodic, the Orthic is a right triangle (Figure 4). If a/b > αh some Orthics are obtuse (a family always contains acutes). 6 Its vertices are the feet of the altitudes. D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 83

Figure 2: Left: If T = ABC is acute (blue), its Orthic T ′ = A′B′C′ is the so-called Fagnano Triangle [26], whose properties include: (i) inscribed triangle of minimum perimeter, (ii) a 3- ′ ′ periodic of T , i.e., the altitudes of T are bisectors of T , i.e., the Orthic Incenter X1 coincides with the Orthocenter X4. Right: If T is obtuse, two of the Orthic’s vertices lie outside T , and ′ X4 is exterior to T . T is the Orthic of both T and acute triangle Te = AX4C. So the Orthic is the latter’s Fagnano Triangle, i.e., B is where both altitudes and bisectors meet. The result is ′ that if T is obtuse at B, the Incenter X1 of the Orthic is B. Video: [23, PL#06]

Figure 3: An a/b > α4 EB is shown (black). Let T and Th be the 3-periodic and its Orthic Triangle (blue and purple, respectively). (a) T is acute (X4 is interior to the EB), and Ih = X4. (b) X4 is on the EB and T is a right triangle. Th degenerates to a segment. (c) X4 is exterior to the EB. Two of Th’s vertices are outside T . Ih is pinned to T ’s obtuse vertex, on the EB. X4 is an Excenter of the 3-periodic. The complete locus of Ih comprises therefore 4 elliptic arcs (thick purple) duck-taped at the corners. Video: [23, PL#07]

Proof. The orthic of an isosceles triangle with vertices A = (a, 0), B = ( u, v) and C = ( u, v) is the isosceles triangle with vertices: − − − A′ = ( u, 0) − 2 2 2 2 u(a + u) + v (2a + u) v((a + u) v ) B′ = − , − (a + u)2 + v2 (a + u)2 + v2) u(a + u)2 + v2(2a + u) v((a + u)2 v2) C′ = − , − (a + u)2 + v2 − (a + u)2 + v2 84 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard

Figure 4: Left: At a/b = αh 1.174, a sideways 3-periodic (blue) has a right-triangle Orthic ≃ ′ (purple). If a/b > α some Orthics in the family are obtuse. Right: At a/b = α 2.605, when h h≃ the 3-periodic is an upright isosceles (obtuse since a/b > αh), its extraverted Orthic is also a right triangle.

It is rectangle, if and only if, B′ A′,C′ A′ =0. This condition is expressed by r(a, u, v)= (a + u)2 v(2a +2u + v)=0h. − − i Using− explicit expression to the 3-periodic vertices [6], obtain that u = u(a, b) = a(δ 2 2 2 2 2 2 2 2 − b )/(a b ) and v = v(a, b) = b √2δ a b /(a b ). So it follows that r(a, u, ) = r(a, b) =− 97a8 52a6b2 122a4b4 + 12a−2b6 +−b8 = 0. The− same argument can be applied≃ to − − the isosceles 3-periodic with vertices: A = (0, b), B =( v(b, a), u(b, a)), C =(v(b, a), u(b, a)) − − − The associated orthic triangle will be rectangle, if and only if, r(b, a)=0. The obtuseness of 3-periodic Orthics is a complex phenomenon with several regimes de- pending on a/b, beyond the scope of this paper. It is explored in [23, PL#08].

2.3. Can a locus be self-intersecting?

The trees are in their autumn beauty, The woodland paths are dry, Under the October twilight the water Mirrors a still sky; Upon the brimming water among the stones Are nine-and-ﬁfty swans. W.B. Yeats

Yeats gave us the idea to look at a curious TC: X59, the “Isogonal Conjugate of X11” [14], i.e., the two centers have reciprocal trilinears. Experimentally, X59 is a continuous curve internally tangent to the EB at its four vertices, and with four self-intersections (Figure 5), and as an animation [23, PL#12]. It intersects a line parallel to and inﬁnitesimally away from either axis on six points, so its degree must be at least 6. Producing analytic expressions for salient aspects of its geometry has proven a tough nut to crack, namely, the following are unsolved: What is the degree of its implicit equation? • What is t in P1(t) = (a cos(t), b sin(t)) such that X59 is on one of the four self- • ◦ intersections? For example, at a/b = 1.3 (resp. 1.5), t, given in degrees is 32.52 (resp. 29.09◦) (Figure 5, left-bottom). ≃ What is a/b such that if X59 is on one of the lower self-intersection on the y-axis, the • ⊥ 3-periodic is a right triangle? Numerically, this occurs when a/b = α59 1.58 and t 27.92◦ (Figure 5, right). ≃ ≃ D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 85

Figure 5: The locus of X59 is a continuous curve with four self-intersections, and at least a sextic. It is tangent to the EB at its four vertices. Top Left: circular EB, X59 is symmetric ◦ about either axis. Bottom Left: a/b = 1.3, at t 32.5 X59 is at the lower self-intersection ≃ ⊥ and the 3-periodic is acute (X4 is interior). Right: at a/b = α59 1.58 the following feat is ≃ possible: X59 is at the lower self-intersection and the 3-periodic is a right-triangle (X4 is on P2). This occurs at t 27.9◦. Video: [23, PL#12] ≃

2.4. Can a locus be non-compact

X26 is the Circumcenter of the Tangential Triangle [30]. Its sides are tangent to the Circum- circle at the vertices. If the 3-periodic is a right-triangle, its hypotenuse is a diameter of the Circumcircle, and X26 is unbounded. We saw above that:

a/b < α4, the 3-periodic family is all-acute, i.e., the locus of X26 is compact (Figure 1, • top left).

a/b = α4, the family is all-acute except when one of its vertices coincides with the top • or bottom vertex of the EB (Figure 1, bottom left). In this case the 3-periodic is a right triangle and X26 is unbounded.

a/b > α4, the family features both acute and obtuse triangles. The transition occurs at • for four right-angle 3-periodics whose X4 is on the EB (Figure 3(b)). Here too X26 flies off to infinity (Figure 6). 86 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard

Figure 6: The locus of X26 for a 3-periodic (blue) in an a/b = 1.25 EB (black). Also shown is the 3-periodic’s Circumcircle (purple) and its Tangential Triangle [30] (dashed green). X26 is the center of the latter’s Circumcircle (solid green). Its locus is non-elliptic. In fact, when a/b α4, the 3-periodic family will contain right-triangles (X4 crosses the EB). At these events, ≥ X26 flies off to infinity. The right inset shows an inversion of X26 with respect to the EB center for various values of a/b. When a/b > α4, the inversion goes through the origin, i.e., X26 is at infinity.

3. Intermezzo: Two unexpected phenomena

3.1. The billiard lays a golden egg

The Bevan Point X40 is known as the Circumcenter of the Excentral Triangle [14]. It is the tangential polygon to the 3-periodic, and can be thought of as its projective dual [17]. We have shown elsewhere the locus of X40 is an ellipse similar to a rotated copy of the D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 87 Billiard. Its semi-axes are given by [8]:

2 2 a40 = c /a, b40 = c /b.

Proposition 5. At a/b = √2 i.e., the top and bottom vertices of the X40 touch the Billiard’s top and bottom vertices.

Proof. This follows from imposing b40 = b. What we got next was unexpected (see Figure 7):

Figure 7: An a/b = ϕ EB is shown golden. Also shown is a sample 3-periodic (blue). The Bevan Point X40 is the Circumcenter of the Excentral Triangle (solid green). At this EB aspect ◦ ratio, the locus of X40 (purple) is a 90 -rotated copy of the EB. The region common to both ellipses is 70.48% the area of either ellipse. Video: [23, PL#13]. ≃

Proposition 6. At a/b = (1+ √5)/2= ϕ, the Golden Ratio, the locus of X40 is identical to a 90◦-rotated copy of the EB

Proof. This follows from imposing b40 = a. Remark 1. Let A be area of the four-corner region common to an ellipse and its 90◦-rotated −1 2 copy. It can be shown A/Aell = 4 csc 1+(a/b) /π, where Aell = πab, is the area of the ellipse. For a/b = ϕ, A/A 0.704833h.p i ell ≃ 3.2. A derived triangle railed onto the EB

′ ′ ′ Consider a 3-periodic’s Anticomplementary Triangle (ACT) [30] and its Intouchpoints i1, i2, i3 (Figure 8). Remarkably:

Theorem 2. The locus of the Anticomplementary Triangle’s Intouchpoints is the EB. 88 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard

Figure 8: A 3-periodic P1P2P3 is shown (blue). Shown also is the Mittenpunkt X9, at the EB ′ ′ ′ center. The 3-periodic’s Anticomplementary Triangle (ACT) P1P2P3 (dashed blue) has sides ′ ′ ′ parallel to the 3-periodic. The latter’s Intouchpoints i1, i2, and i3 are the feet of perpendiculars dropped from the ACT’s Incenter (X8) to each side (dashed green). The ACT’s Incircle (green) and 9-point circle (the 3-periodic’s Circumcircle, pink) meet at X100, the ACT’s Feuerbach Point. Its locus is also the EB. The Caustic is shown brown. On it there lie X11 and the three Extouchpoints e1, e2, e3. Video: [23, PL#09]

Proof. Consider an elementary triangle with vertices Q1 = (0, 0), Q2 = (1, 0) and Q3 =(u, v). Its sides are s1 = Q3 Q2 , s2 = Q3 Q1 , and s3 =1. | − | | − | Let E be its Circumbilliard, i.e., the Circumellipse for which Q1Q2Q3 is a 3-periodic EB trajectory. The following implicit equation for E was derived [6]: 2 2 2 2 E(x, y)= v x +(u +(s1 + s2 1)u s2)y + − − 2 v(1 s1 s2 2u)xy + v(s2 + u)y v x =0 − − − − ′ ′ ′ The vertices of the ACT are given by Q1 =(u 1, v), Q2 =(u +1, v), Q3 = (1 u, v), and its Incenter7 is: − − − 2 ′ s2(s1 1) + (1 s1 + s2)u u X1 = s1 s2 + u, − − − . − v ′ The ACT Intouchpoints are the feet of perpendiculars dropped from X1 onto each side of the ACT, and can be derived as:

′ s1(u 1)u + s2 v(s1 1) i1 = − , − , s2 s2

′ (u 1)(s2 1) (s2 1)v i2 = − − , − , s2 s2 ′ i3 = [s1 s2 + u, v] . − ′ ′ ′ Direct calculations shows that E(i1) = E(i2) = E(i3)=0. Besides always being on the EB, ′ ′ ′ the locus of the Intouchpoints cover it. Let P1(t)P2(t)P3(t) be a 3-periodic and P1(t)P2(t)P3(t) ′ ′ ′ its ACT. For all t the Intouchpoint i1(t) is located on the side P2(t)P3(t) of the ACT and on the elliptic arc arc(P1(t)P3(t)) (Figure 8). Therefore, when P1(t) completes a circuit on the ′ ′ ′ EB, i1(t) will have to complete a similar tour. Analogously for i2(t) and i3(t).

7 This is the Nagel Point X8 of the original triangle. D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 89

′ Figure 9: As one endpoint P1 of a billiard trajectory is slid CCW to P1, its tangency point C with the Caustic (brown) slides in the same direction to C′. This must be the case since ′ ′ ′ ′ ∗ P1P2 corresponds to a CCW rotation about P1 of segment P1P2 (pink) parallel to P1P2 (see pink arrow). By convexity, said rotation will ﬁrst touch the Caustic at C′, lying “ahead” of C. Repeating this for the P2P3 segment of a 3-periodic (not shown), it follows said vertices will move in the direction of P1.

4. Second movement: motion

Let P1,P2,P3 be the vertices of a 3-periodic (Appendix B).

Proposition 7. If P1 is slid along the EB in some direction, P2 and P3 will slide in the same direction.

Proof. Consider the tangency point C of P1P2 with the confocal Caustic (Figure 9). Since this segment remains tangent to the Caustic for any choice of P1(t), a counterclockwise motion of P1(t) will cause C to slide along the Caustic in the same direction, and therefore P2(t) will do the same. The simultaneous monotonic motion of 3-periodic vertices is shown in Figure 10, note the non-linear progress of P2,P3. Alternatively, we could have linearized their motion using the

Figure 10: As P1(t) moves monotonically forward, so do P2(t) and P3(t), albeit with varying velocities with respect to t. In the text we mention an alternate parametrization (Poritsky- Lazutkin) under which the three lines would become straight. 90 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard so-called Poritsky-Lazutkin string length parameter η for C on the Caustic (Figure 9) given by [19, 16, 11]: dη = κ2/3 ds, where s is the arc length along the Caustic, and κ is the curvature. Both η and s are related to the parameter t on the billiard by elliptic functions. Adjusting conveniently with a constant factor, one has η η +1 and for any η the other vertices correspond to η +1/3 and η +2/3. ≡ o o o 4.1. Non-monotonicity: a ﬁrst brushing Let α =2 2/5 1.2649. act ≃ Propositionp 8. The motion of the ACT Intouchpoints is as follows: a/b < α : monotonic in the direction of P1(t). • act a/b = α : monotonic in the direction of P1(t), except for two instantaneous stops when • act passing at EB top and bottom vertices.

a/b > αact: non-monotonic with four reversals of velocity, a ﬁrst (resp. second) pair of • reversals near the EB’s top (resp. bottom) vertex (Figure 8).

Proof. As before, let a 3-periodic P1P2P3 be parametrized by a leading vertex

P1(t)=(x1,y1)=(a cos t, b sin t). ′ Its ACT Pi is given by double-length reﬂections of Pi about the Barycenter X2 [30]. Taking the ACT as the reference triangle, use Intouchpoint Trilinears 0 : s1s3/(s1 s2 +s3) :: [30, Contact ′ − ′ 1 1 = π Triangle] to compute an Intouchpoint i1(t)=(x (t),y (t)) it follows that x1(t) t 2 = 0 is equivalent to 5a2 8b2 =0. This yields the result. | − See [22, PL#09] for an animation of the non-monotonic case. As we had been observ- ing the EKG-like graph in Figure 11 (right), we stumbled upon an unexpected property, namely, the ﬁxed linear relation between a 3-periodic vertex and its corresponding (opposite) Extouchpoint: ′ ′ Proposition 9. Let Pi(t)=(xi,yi) be one of the 3-periodic vertices and ei = (xi,yi) be its 8 9 corresponding Extouchpoint on the Caustic, where ac, bc are the latter’s semi-axes. Then , for all t: ′ a yi ac yi = ′ (2) b xi bc xi Equivalently, for e (t′) = [a cos t′, b sin t′], then tan(t) = tan(t′), i.e., t′ = t π. i c c ± Proof. This property follows directly10 from [17, Lemma 3]. In fact, more general properties of the “Poncelet grid” are described in the reference above. The result reported here is a particular case and can also be demonstrated by simplifying rather long symbolic parametrics with a Computer Algebra System (CAS). Furthermore, since a =(δ b2)a/c2 and b =(a2 δ)b/c2 [9]: c − c − y δ b2 y′ = − (3) x a2 δ x′ − 8 Where Pi−1(t)Pi+1(t) touch the Caustic. 9 ′ ′ It can be shown (2) also holds if x ,y are the coordinates of an Excenter and ac,bc are the semi-axes of the excentral locus, known to be an ellipse [6]. 10 We thank A. Akopyan for pointing this out. D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 91

Figure 11: The “EKG” of ACT motion, to be interpreted as a ﬂat torus. On the horizontal axis is parameter t of P1(t) = [a cos(t), b sin(t)], with a/b = 1.5. On the vertical is the t parameter ′ for P1, X11, X100 and an ACT Intouchpoint i1. These all lie on the EB and are shown blue, red, dashed red, and green, respectively. Also shown is Extouchpoint e1 on the Caustic (dashed ′ ′ ′ blue). Just for it, the vertical axis represents t in e1 = [ac cos(t ), bc sin(t )], where ac, bc are the Caustic semi-axes. By Proposition 9, t′ = t π. Notice the only non-monotonic motion is that ′ ± ′ of i1 since a/b > αact 1.265. To see this, a vel(i1) of its velocity is also shown (dashed green), ≃ ′ containing two negative regions corresponding to 4 critical points of position. For vel(i1) ignore units and the fact that values near 0◦, 180◦ are not shown, these are all positive and above the vertical scale.

4.2. A non-monotonic triangle center Dovetailing into the non-monotonicity of the ACT’s Intouchpoints is a similar behavior by X88, the Isogonal Conjugate of X44, known to lie on the EB and to be collinear with X1 and X100 [14]. The latter is veriﬁed by the vanishing determinant of the 3 3 matrix whose rows × are the trilinears of X1,X100,X88 [30, under “Collinear”, eqn 9]:

1 1 1  1 1 1  det 0 s2 s3 s3 s1 s1 s2 ≡  −1 −1 −1   s2 + s3 2 s1 s1 + s3 2 s2 s1 + s2 2 s3   − − −  Furthermore X100 is a very special point: it lies on the EB and on the Circumcircle simulta- neously [14]. Let α88 =( 6+2√2)/2 1.485. p ≃ Proposition 10. At a/b = α88, the y velocity of X88 vanishes when the 3-periodic is a sideways isosceles.

Proof. Parametrize P1(t) in the usual way. At t = 0, P1 = (a, 0) it can be easily checked ′ that X88 = ( a, 0). Solve y88(t) t=0 = 0 for a/b. After some algebraic manipulation, this is − 4 2 | equivalent to solving 4x 12x +7=0, whose positive roots are ( 6 2√2)/2. α88 is the largest of the two. − p ±

As shown in Table 1, there are three types of X88 motion with respect to P1(t): monotonic, with stops at the EB vertices, and non-monotonic. An equivalent statement is that the line family X1X100 is instantaneously tangent to its envelope [30] at X88. Figure 12 shows that said envelope lies (i) entirely inside, (ii) touches at 92 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard

Table 1: Conditions for the type of motion of X88 with respect to P1(t). Video: [23, PL#11]

a/b vs. α88 motion comment < CW monotonic = CW stops at EB vertices > CW+CCW non-monotonic

vertices of, or (iii) is partially outside, the EB, when a/b is less than, equal, or greater than α88, respectively. Each such case implies the motion of X88 is (i) monotonically opposite to P1(t), (ii) opposite but with stops at the EB vertices, or (iii) is non-monotonic. The reader is challenged to ﬁnd an expression for parameter t in P1(t) where the motion

Figure 12: Collinear points X1, X100, X88 shown for billiards with a/b less than (top-left), equal (top-right), or greater (bottom) than α88 1.486, respectively. In each such case, the motion ≃ of X88 relative to the 3-periodic vertices will be monotonic, with stops at the vertices, or non- monotonic, respectively. Equivalently, the motion of X88 is opposite to P1, stationary, or in the direction of P1 if the instantaneous center of rotation E of line X1X100 lies inside, on, or outside the EB. The locus of E (the envelope of X1X100) is shown purple. Notice it only “pierces” the EB when a/b > α88 (bottom), i.e., only in this case can the motion of X88 be non-monotonic. Video: [23, PL#12] D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 93

of X88 changes direction. The following additional facts are also true for X88:

Proposition 11. X88 coincides with a 3-periodic vertex if and only if s2 = (s1 + s3)/2. In this case, X1 is the midpoint between X100 and X88 [12] (Figure 13, bottom left).

Proof. The ﬁrst trilinear coordinate of X88 is 1/(s2 + s3 2s1) [14], and of a vertex is 0. − Equating the two yields s2 = (s1 + s3)/2. Consider a triangle of reference P1 = ( 1, 0), 2 2 2 2 − P2 =(u, v), P3 = (1, 0). Its circumcircle is given by v(x + y )+(1 u v )y v =0. Under 2 − − − the hypothesis s2 = (s1 + s3)/2 it follows that v = √12 3u /2, s1 = 2 u/2, s2 = 2 and − − s3 =2+ u/2. Therefore the incenter is

2 I =(s1P1 + s2P2 + s3P3)/(s1 + s2 + s3)=(u/2, √12 3u /6). − The intersection of the straight line passing through P2 and I with the circumcircle of the triangle o reference is the point D = (0, √12 3u2/6). Therefore, I is the midpoint of B and − 2 D = X100. Moreover, P1 D = P2 D = I D = √48 3u /6 . | − | | − | | − | − It is well-known that the only right-triangle with one side equal to the average of the ⊥ other two is 3:4:5. Let α88 =(7+ √5)√11/22 1.3924. Referring to Figure 13 (right): ≃ Proposition 12. The only EB which can contain a 3:4:5 3-periodic has an aspect ratio ⊥ a/b = α88.

Figure 13: X88 is always on the EB and collinear with X1 and X100 [14]. Let ρ (shown above each picture) be the ratio X1 X100 / X1 X88 . Top Left: The particular 3-periodic shown is | − | | − | obtuse (X4 is exterior), and ρ> 1, i.e., X1 is closer to X88. Bottom Left: When X88 coincides with a vertex, if sidelengths are ordered as s1 s2 s3, then s2 = (s1 + s3)/2, and X1 becomes ≤ ≤ ⊥ the midpoint of X88X100, i.e., ρ = 1. Right: If a/b = α88 1.39, when X88 is on a vertex, the ≃ 3-periodic is a 3:4:5 triangle (X4 lies on an alternate vertex). Video: [23, PL#11] 94 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard

Proof. With a/b > α4 = 2 √2 1 1.352 the 3-periodic family contains obtuse triangles amongst which there alwaysp are− 4 right≃ triangles (identical up to rotation and reﬂection). Consider the elementary triangle P1 = (0, 0), P2 = (s1, 0), and P3 = (s1,s2) choosing s1,s2 2 2 integers such that s3 = s3 = s1 + s2 is an integer. The Circumbilliard [9] is given by:

2 p 2 E9(x, y)= s2x +(s3 s1 s2)xy + s1y s1s2x s1(s1 s3)y =0. − − − − − Squaring the ratio of the Eigenvalues of E9’s Hessian yields the following expression for a/b:

s1 + s2 + s3 (3 s3 2 s1 2 s2) [a/b](s1,s2,s3)= − − (s1 + sp2 +3 s3)(s1 + s2 s3) − p Table 2 shows a/b for the ﬁrst 5 Pythagorean triples ordered by hypotenuse11.

Table 2: First 5 Pythagorean triples ordered by hypotenuse. a/b is the aspect ratio. of the EB which produces 4 triangles homothetic to the triple.

(s1,s2,s3) a/b a/b ≃ 3, 4, 5 (7 + √5)√11/22 1.392 5, 12, 13 √14(√65+ 17)/56 1.674 8, 15, 17 √111(√85+ 23)/222 1.529 7, 24, 25 √159(5 √13 + 31)/318 1.944 20, 21, 29 √6(√145 + 41)/96 1.353

4.3. Swan Lake

But now they drift on the still water, Mysterious, beautiful; Among what rushes will they build, By what lake’s edge or pool Delight men’s eyes when I awake some day To ﬁnd they have ﬂown away? W.B. Yeats

In addition to X88 and X100, a gaggle of 50+ other TCs are also known to lie on the EB [14, X(9)], e.g., X162, X190, X651, etc. We have found experimentally that the motion of X100 and X190 is always monotonic. 8 6 4 In contradistinction: Let α162 1.1639 be the only positive root of 5x +3x 32x + 52x2 36. ≃ − − Proposition 13. The motion of X162 with respect to P1(t) is non-monotonic if a/b > α162.

Proof. The trilinear coordinates of X162 are given by 1 1 1 2 2 2 2 2 : 2 2 2 2 2 : 2 2 2 2 2 (s2 s3)(s2 + s3 s1 (s3 s1)(s3 + s1 s2) (s1 s2)(s1 + s3 s3) · − − − − − − 11 The a/b which produces 3:4:5 was ﬁrst computed in connection with X88 [12]. D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 95 v

Figure 14: Level curves F = 0 (red) and G = 0 (black).

Let a 3-periodic P1P2P3 be parametrized by P1(t)=(a cos t, b sin t), with P2(t) and P3(t) given in Appendix B. Using the trilinear coordinates above, we have X162(t)=(x126(t), y126(t)) at π π t = 2 , P1 = (0, b) and X162( 2 )=(0, b). ′ = π Solve x162(t) t 2 = 0 for a/b. After some long algebraic symbolic manipulation, this | 8 6 4 2 is equivalent to solving 5x +3x 32x + 52x 36 = 0, whose positive roots is α162 1.16369. − − ≃

Since α88 >α162, setting a/b > α88 implies both centers will move non-monotonically. If the EB be a lake, their joint motion is a dance along its margins. Over a complete revolution of P1(t) around the EB, X88 and X162 wind thrice around it, however:

Proposition 14. X88 and X162 never coincide, therefore their paths never cross each other.

Proof. Consider an elementary triangle P1 = ( 1, 0), P2 = (1, 0) and P3 = (u, v). Obtain − cartesian coordinates for X88 and X162 using their trilinears (Propositions 11 and 13). The equation X88 = X162 is given by two algebraic equations F (u,v,s1,s2) = G(u,v,s1,s2)=0 2 2 2 2 of degree 17 with s1 = (u 1) + v = P3 P2 and s2 = (u + 1) + v = P2 P1 . − | − | | − | Particular solutions of thesep equations are equilateral triangles withp P3 = (0, √3). Analytic ± and graphic analysis reveals that the level curves F = G =0 are as shown in Figure 14. Therefore the equilateral triangle is the only one such that the triangle centers X88 and X162 are equal. It is well known that an equilateral billiard orbit occurs only when the billiard ellipse is a circle. This ends the proof. Their never-crossing joint motion as well as the instants when they come closest is illustrated in Figure 15. These paths can also be viewed as non-intersecting loops on the torus shown in Figure 16. An animation of this intricate motion can be viewed in [23, PL#14] with triangle centers imagined as swans. Though we lack a theory for non-monotonicity of EB-bound TCs Xi, we think it is ruled by at least the following aspects: When the 3-periodic is an isosceles, will X lie at the summit vertex or below the base? • i As the 3-periodic rotates about the EB, does X follow it or move counter to it? • i Can X be non-monotonic? For example, neither X100 nor X190 ever are. If so, what is • i the aspect ratio αi which triggers it?

Still murky is how the above derive from the Trilinears which specify Xi. We refer the reader to an animation depicting the joint motion of 20-odd such centers [23, PL#15]. 96 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard

′ Figure 15: Top: location t of 3-periodic vertices P1 (blue), P2 (dashed blue), and P3 (dashed blue), as well as X88 (red), and X162, plotted against the t parameter of P1(t). Bottom: absolute parameter diﬀerence along the EB between X88 and X162. Notice there are 12 identical maxima at 180◦ occurring when the two centers at the left and right vertices of the EB. Additionally, there are 12 identical minima whose values can be obtained numerically. The fact that the minimum is above zero implies the points never cross. Note the highlighted minimum at t 41◦: ≃ it is referred to in Figure 16.

Figure 16: Visualizing the joint motion of P1(t), X88, X162 on the surface of a torus. The meridians (circles around the smaller radius) correspond to a given t (a solid black meridian is wound at t = 0. The parallels represent a ﬁxed location on the Billiard boundary. The curves for X88 and X162 are thrice-winding along the torus though never intersecting. To be determined: an analytic value for t where X88 and X162 are closest (there are 12 solutions). The dashed meridian represents one such minimum which for a/b = 2 occurs at t 41◦. Notice it does not ≃ coincide with any critical points of motion. D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 97 4.4. Summary of phenomena Tables 3 and 4 respectively summarize this section’s various loci phenomena, and notable a/b thresholds. Table 3: Loci phenomena for various Triangle Centers.

Point Type of Locus Non-Monotonic Comments X11 Caustic – reverse dir. Extouchpts. Caustic – forward dir. X4 Upright Ellipse – Inside EB when a<α4 Orthic X1 Can be 4-piece Ell. – If a<α4 is X4 locus, else 4-piecewise ellipse: 2 arcs from X4 locus, 2 arcs from EB X26 Can be non-compact – When a/b α4, goes to ≥ ∞ if 3-periodic is right triangle X40 Upright Ellipse – At a/b = ϕ identical to EB X59 At Least Sextic – 4 Self-Intersections ⊥ X88 EB a/b > α88 at a/b = α88 >α4 contains 3:4:5 triangle X162 EB a/b > α162 never crosses X88

ACT Intouchpts. EB Boundary a/b > αact

Table 4: Aspect ratio thresholds, the degree of the polynomial used to ﬁnd them, and their eﬀects on loci phenomena.

symbol a/b Degree Signiﬁcance α162 1.164 8 above it, motion of X162 is non-monotonic

αh 1.174 8 above it, some 3-periodic Orthics can be obtuse.

αact 1.265 2 above it, motion of ACT Intouchpoints is non-monotonic α4 1.352 4 locus of X4 is tangent to EB; above it, some 3-periodics are obtuse ⊥ α88 1.392 4 with X88 on a vertex, sidelengths are 3:4:5 α88 1.486 4 above it, motion of X88 is non-monotonic ∗ α4 1.510 6 X4 locus identical to rotated EB ⊥ α59 1.580 – when X59 is at self-intersection, 3-periodic is right triangle ϕ 1.618 2 X40 locus identical to rotated EB 98 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 5. Conclusion

Examining the cygnine dynamic geometry and loci of 3-periodics under the prism of Classical Triangle Geometry has been productive. Surely additional secrets still hide under the EB’s plumage. However, this haphazard experimental approach needs theoretical teeth. Toward that end, we submit the following questions to the reader: Can a Triangle Center be found such that its locus can intersect a straight line more • than 6 times? The questions about X59 in Section 2.3. • What causes a Triangle center to move monotonically (or not), forward or backward, • with respect to the the monotonic motion of 3-periodic vertices? What is t in P1(t) at which point X88 or X162 reverses motion? When do they come • closest? What can be said about the joint motion of other pairs in the 50+ list of EB-bound • points provided in [14, X(9)]? Which are monotonic, which are not, is there a Pavlova or Baryshnikov amongst them? With [20] one observes X823 reverses direction at the exact moment a 3-periodic vertex • crosses it. Can this be proven? Videos mentioned herein are on a playlist [23], with links provided on Table 5. The reader is especially encouraged to interact with some of the above phenomena using our online applet [20]. A gallery of loci generated by X1 to X100 (as well as vertices of derived triangles) is provided in [21].

Table 5: Videos mentioned in the paper. Column “PL#” indicates the entry within the playlist [23].

PL# Video Title Section 01 X9 stationary at EB center 1 02 Loci for X1 ...X5 are ellipses 1 03 Elliptic locus of Excenters similar to rotated X1 1 04 Loci of X11, X100 and Extouchpoints are the EB 1 05 Non-Ell. Loci of Medial, Intouch and Feuerbach Vertices 1 06 Pinning of Incenter of Orthic to Obtuse Vertex 2.1 07 Locus of Orthic Incenter is 4-piece ellipse 2.2 08 Locus of X4, Orthic X1,X4, and Orthic’s Orthic X1 2.2 09 Non-monotonic motion on the EB of Anticompl. Intouchpoints 4.1 10 Locus of X88 is on the EB and can be non-monotonic 4.2 11 Non-monotonic motion of X88 and X1X100 envelope 4.2 12 Locus of X59 with 4 self-intersections 2.3 13 Locus of X40 and the Golden Billiard 3.1 14 Swan Lake: the dance of X88 and X162 4.3 15 Peter Moses’ Points on the EB 4.3 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 99 Acknowledgements

Warm thanks go out to Clark Kimberling, PeterMoses, Olga Romaskevich, and Mark Helman for helpful discussions. The second autor is fellow of CNPq and coordinator of Project PRONEX/CNPq/FAPEG 2017 10 26 7000 508.

References

[1] A. Akopyan, R. Schwartz, S. Tabachnikov: Billiards in ellipses revisited. URL https://arxiv.org/abs/2001.02934 (2020). [2] M. Bialy, S. Tabachnikov: Dan Reznik’s identities and more. URL https://arxiv.org/abs/2001.08469 (2020). [3] H.S.M. Coxeter, S.L. Greitzer: Geometry Revisited. New Mathematical Library, vol. 19, Random House, Inc., New York 1967. [4] V. Dragović, M. Radnović: Caustics of Poncelet polygons and classical extremal polynomials. Regul. Chaotic Dyn. 24(1), 1–35 (2019). URL https://doi.org/10.1134/S1560354719010015 [5] C. Fierobe: On the circumcenters of triangular orbits in elliptic billiard. URL https://arxiv.org/pdf/1807.11903.pdf (2018). Submitted [6] R. Garcia: Elliptic billiards and ellipses associated to the 3-periodic orbits. Amer. Math. Monthly 126(06), 491–504 (2019). URL https://doi.org/10.1080/00029890.2019.1593087 [7] R. Garcia, D. Reznik: Similar families: Poristic triangles and 3-periodics in the elliptic billiard. URL https://arxiv.org/abs/2004.13509 (2020). [8] R. Garcia, D. Reznik, J. Koiller: Loci of 3-periodics in an elliptic billiard: why so many ellipses? URL https://arxiv.org/abs/2001.08041 (2020). [9] R. Garcia, D. Reznik, J. Koiller: New properties of triangular orbits in elliptic billiards. URL https://arxiv.org/abs/2001.08054 (2020). [10] G. Glaeser, H. Stachel, B. Odehnal: The universe of conics. From the an- cient Greeks to 21st century developments. Springer Spektrum, Berlin 2016. URL https://doi.org/10.1007/978-3-662-45450-3. [11] A. Glutsyuk: On curves with poritsky property. arXiv (2019). [12] M. Helman: Locus of X88 and aspect ratio for 3:4:5 3-periodic. Private Communi- cation (January 2020). [13] V. Kaloshin, A. Sorrentino: On the integrability of Birkhoff billiards. Phil. Trans. R. Soc. A(376) (2018). DOI https://doi.org/10.1098/rsta.2017.0419 [14] C. Kimberling: Encyclopedia of triangle centers. URL https://faculty.evansville.edu/ck6/encyclopedia/ETC.html (2019). [15] C. Kimberling: Polynomial triangle centers on the line at infinity. J. Geom. 111(10) (2020). URL 10.1007/s00022-020-0522-y [16] V. Lazutkin: The existence of caustics for a billiard problem in a convex domain. Math. USSR Izvestija 7, 185–214 (1973). [17] M. Levi, S. Tabachnikov: The Poncelet grid and billiards in ellipses. Amer. Math. Monthly 114(10), 895–908 (2007). URL https://doi.org/10.1080/00029890.2007.11920482 100 D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard [18] B. Odehnal: Poristic loci of triangle centers. J. Geometry Graphics 15(1), 45–67 (2011). [19] H. Poritsky: The billiard ball problem on a table with a convex boundary – an illustra- tive dynamical problem. Ann. of Math. 51(2), 446–470 (1950). [20] D. Reznik: Applet showing the locus of several triangular centers. URL https://editor.p5js.org/dreznik/full/i1Lin7lt7 (2019). [21] D. Reznik: Loci of triangle centers x(1)–x(100): Orbits and 6 derived triangles. URL https://bit.ly/2TCsHFU (2019). [22] D. Reznik: YouTube playlist for mathematical intelligencer. URL https://bit.ly/2kTvPPr (2019). [23] D. Reznik: Playlist for “Loci of Triangular Orbits in an Elliptic Billiard: Intriguing Phenomena”. URL https://bit.ly/2vvJ9hW (2020). [24] D. Reznik, R. Garcia, J. Koiller: Can the elliptic billiard still surprise us? Math. Intelligencer 42 (2019). URL https://rdcu.be/b2cg1 [25] O. Romaskevich: On the incenters of triangular orbits on elliptic billiards. Enseign. Math. 60(3-4), 247–255 (2014). URL https://arxiv.org/pdf/1304.7588.pdf [26] U.A. Rozikov: An Introduction To Mathematical Billiards. World Scientific Publishing Company, 2018. [27] S. Tabachnikov: Geometry and Billiards. Student Mathematical Library vol. 30, American Mathematical Society, Providence, RI 2005. URL http://www.personal.psu.edu/sot2/books/billiardsgeometry.pdf [28] S. Tabachnikov: Projective configuration theorems: old wine into new wineskins. In: S. Dani, A. Papadopoulos (eds.): [Geometry in History, Springer Verlag 2019, pp. 401–434. URL https://arxiv.org/pdf/1607.04758.pdf [29] J.H. Weaver: Invariants of a poristic system of triangles. Bull. Amer. Math. Soc. 33(2), 235–240 (1927). URL https://doi.org/10.1090/S0002-9904-1927-04367-1 [30] E. Weisstein: Mathworld. URL http://mathworld.wolfram.com (2019). [31] P. Yiu: Conic construction of a triangle from its incenter, nine-point center, and a vertex. J. Geometry Graphics 16(2), 171–183 (2012).

A. Obtuse triangles and their orthics

′ ′ ′ ′ ′ Let T = ABC be any triangle and T = A B C its Orthic (Figure 2). Let X1 be the orthic’s Incenter.

′ ′ Lemma 2. If T is acute, the Incenter X1 of T is the Orthocenter X4 of T .

Proof. This is Fagnano’s Problem, i.e., the Orthic is the inscribed triangle of minimum perimeter, and the altitudes of T are its bisectors [26, Section 3.3]. Since the altitudes or T are bisectors of T ′ this completes the proof.

′ ′ Lemma (1). If T is obtuse, the Incenter X1 of T is the vertex of T subtending the obtuse angle. D. Reznik, R. Garcia, J. Koiller: The Ballet of Triangle Centers on the Elliptic Billiard 101 Proof. Let B be the obtuse angle. Then B′ will be on the longest side of T whereas A′ (resp. ′ ′ ′ C ) will lie on extensions of BC (resp. AB), i.e., A and C are exterior to T . Therefore, X4 will be where altitudes AA′ and CC′ meet, also exterior to T . Since AA′ CB and CC′ AB, ′ ′ ⊥ ⊥ then CA and AC are altitudes of triangle Te = AX4C. Since these meet at B, the latter ′ ′ ′ ′ ′ is the Orthocenter of Te and T = A B C is its Orthic, noting that the pre-image of T ′ ′ ′ ′ comprises both T and Te. By Lemma 2, lines CA , AC ,X4B are bisectors of T , therefore their meetpoint B is the Incenter of T ′.

′ Corollary 1. When T is obtuse, Te = AX4C is acute and the Excentral Triangle of T . ′ Since all vertices of T lie on the sides of Te, this is the situation of Lemma 2, i.e., Te is ′ acute. Notice the sides of Te graze each vertex of T perpendicular to the bisectors, which is the construction of the Excentral Triangle. Let Q be a generic triangle and Qe its Excentral [30]. Let θi be angles of Q and φi those π − θi of Qe opposite to the θi’s. By inspection, φi = 2 , i.e., all excentral angles are less than π/2. ′ Corollary 2. Cor.2 If T is obtuse, X4 is an Excenter of T [3].

X4 is the intermediate vertex of Te.

B. 3-Periodic vertices

Results here were first derived in [6]. The EB semi-axes are a>b> 0. Let Pi = (xi,yi) denote the 3-periodic vertices, i =1, 2, 3, and α denote the angle of incidence (and reflection) with respect to the Elliptic Billiard normal at P1. This is given by: a2b a2 b2 + 2 √a4 b2c2 cos α = p− 2 − 4 2 2 − , c a c x1 − 2 2 p p2x p2y where c = √a b . Parametrize P1 = (a cos t, b sin t). Coordinates x2 = and y2 = − q2 q2 are given by: 4 2 2 2 2 3 6 3 p2x = b a + b cos α a x1 2a cos α sin αy1 − 4 2 2 2 − 2 − 2 4 2 2 + a (a 3 b ) cos α + b x1 y1 2 a b cos α sin αx1y1, 6 − 3 4 2 2 −2 2 3 p2y = 2b cos α sin αx1 a a + b cos α b y1 2 4 − 2 4 2 2− 2 2 2 + 2 a b cos α sin αx1y1 + b (b 3 a ) cos α + a x1y1 4 2 2 2 2 2 4 −2 2 2 2 2 q2 = b a (a b ) cos α x1 +a b + (a b ) cos α y1 2−2 2 − 2 − 2 a b a b cos α sin αx1 y1. − − p3x p3y Likewise, x3 = and y3 = are given by: q3 q3 4 2 2 2 2 3 6 3 p3x = b a b + a cos αx1 + 2 a cos α sin αy1 4 − 2 2 2 2 2 4 2 2 + a cos α a 3 b + b x1 y1 + 2 a b cos α sin αx1y1 6 −3 4 2 2 2 2 3 p3y = 2 bcos α sin αx1 + a b b + a cos α y1 − 2 4 2 4− 2 2 2 2 2 2 a b cos α sin αx1y1 + b a + b 3 a (cos α) x1 y1, − − 4 2 2 2 2 2 4 2 2 2 2 2 q3 = b a a b cos α x1 + a b + a b cos α y1 2−2 2 − 2 − + 2 a b a b cos α sin αx1 y1. − Received March 24, 2020; final form May 5, 2020